# Increasing Exam Expected Mark

You are in an exam, and are presented with the following question:

Write down what mark do you expect to take in this exam... If you get it right in range of +/-10 % then you will take 10% bonus... if wrong (or not answered) you will lose 5%

Assume that you have no idea of how you are going to perform in this exam. How would you choose a mark that maximizes your expected return mark?

In other words, I need to help deriving an equation for the expected mark given a decided mark, if that is possible...

• I interpreted "10% bonus" as an additional 10% towards your return mark. For example, 30% goes to 40% with this bonus. However, it could be that your return mark is increased by 10% (30% goes to 33%). Would you clarify? Jan 14 '11 at 15:25
• 30% goes to 40%.... Jan 14 '11 at 16:58

First a couple of assumptions: 1. All marks are equally likely. 1. If you guess your mark to be 95 and you get 95, your return mark is 100 not 105. 1. Similarly, if your exam mark is 1 and you guess 50 (say), then your return mark is 0 not -4. 1. I'm only considering discrete marks, that is, values 0, ..., 100.

Suppose your guessed mark is $g=50$. Then your expected return mark is: $$\frac{\sum_{i=0}^{34} + \sum_{i=50}^{70} + \sum_{i=56}^{95}}{101} = 49.15482$$ This is for a particularly $g$. We need to repeat this for all $g$. Using the R code at the end, we get the following plot: Since all marks are equally likely you get a plateau with edge effects. If you really have absolutely no idea of what mark you will get, then a sensible strategy would be to maximise the chance of passing the exam. If the pass mark is 40%, then set your guess mark at 35%. This now means that to pass the exam, you only need to get above 35% but more importantly your strategy for sitting the exam is to answer every question to the best of your ability.

If your guess mark was 30%, and towards the end of the exam you thought that you would score 42%, you are now in the strange position of deciding whether to intentionally make an error (as 42% results in a return mark of 37%).

Note: I think in most real life situations you would have some idea of how you would get on. For example, do you really think that you have equal probability of getting between 0-10%, 11-20%, ..., 90-100% in your exam.

R code

f =function(s) {
mark = 0
for(i in 0:100){
if(i < (s-10) | i > (s + 10))
mark = mark +  max(0, i-5)
else
mark = mark + min(i+10, 100)
}
return(mark/101)
}
s = 0:100
y = sapply(s, f)
plot(s, y)

• With a uniform prior distribution wouldn't guessing 30 be equivalent to guessing 35, since both will have 65 actual test grades that will result in a passing grade? Jan 15 '11 at 16:12
• @Andy-w Good point, I should have explained my reasoning. I've now updated my answer. Jan 15 '11 at 21:43

I'm not sure if this would be a funny game or your professor is mildly sadistic. It would be torturous for students who are right on the edge of passing (which we may expect them to be the worst guessers!) Sorry not an answer but I couldn't help myself. Use the bootstrap! Take lots of practice exams and estimate what your score will be on the real exam. If it does not improve your estimate, it will probably be good preparation!

• +1, I think this is a better strategy if your goal is to maximize the grade and you have prior distribution of test grades. It seems a bit overkill if your only goal is pass the exam though. Jan 15 '11 at 16:14